Suppose the group \(G\) acts on the non-empty set \(A\). Assume this action is faithful (\(\ker(G \to \operatorname{Sym}(A)) = \{1\}\)) and transitive (for all \(a, b \in A\), there exists \(g \in G\) such that \(g \cdot a = b\)).
For \(a \in A\), let \(G_a = \operatorname{Stab}_G(a)\) denote the stabilizer of \(a\). Prove that: \[ \bigcap_{\sigma \in G} \sigma G_a \sigma^{-1} = \{1\}. \]
Suppose that \(G\) is abelian. Prove that \(|G| = |A|\), and deduce that every abelian transitive subgroup of the symmetric group \(S_n\) has order exactly \(n\).
Solution
Goal: Prove that the kernel of a transitive action is the normal core \(\bigcap_{g \in G} g G_a g^{-1} = \{1\}\), and show that transitive faithful abelian actions are regular (\(G_a = \{1\}\) and \(|G| = |A|\)).
1. Part (a): Proof that the Normal Core \(\bigcap_{\sigma \in G} \sigma G_a \sigma^{-1} = \{1\}\):Proof:
1.1. The action of \(G\) on \(A\) is given by a group homomorphism \(\rho: G \to \operatorname{Sym}(A)\), where \(\rho(g)(x) = g \cdot x\).
1.2. The kernel of the action is defined as:\[\ker(\rho) = \{ g \in G \mid g \cdot x = x \text{ for all } x \in A \} = \bigcap_{x \in A} G_x.\]
1.3. Since the action is transitive, every element \(x \in A\) can be written as \(x = \sigma \cdot a\) for some \(\sigma \in G\).
1.4. The stabilizer of \(x = \sigma \cdot a\) is the conjugate subgroup:\[G_x = G_{\sigma \cdot a} = \sigma G_a \sigma^{-1}.\]
1.5. Therefore, the intersection of all stabilizers over all points in \(A\) is:\[\bigcap_{x \in A} G_x = \bigcap_{\sigma \in G} \sigma G_a \sigma^{-1}.\]
1.6. Since the action is faithful by hypothesis, \(\ker(\rho) = \{1\}\).
1.7. Thus:\[\bigcap_{\sigma \in G} \sigma G_a \sigma^{-1} = \ker(\rho) = \{1\}.\]
2. Part (b): Faithful Transitive Actions of Abelian Groups are Regular:Proof:
2.1. Now assume that \(G\) is abelian.
2.2. Since \(G\) is abelian, every subgroup of \(G\) is normal.
2.3. In particular, the stabilizer subgroup \(G_a\) is normal in \(G\):\[\sigma G_a \sigma^{-1} = G_a \quad \text{for all } \sigma \in G.\]
2.4. Substituting this into the intersection formula from Part (a):\[\bigcap_{\sigma \in G} \sigma G_a \sigma^{-1} = \bigcap_{\sigma \in G} G_a = G_a.\]
2.5. By Part (a), this intersection is trivial:\[G_a = \{1\}.\]
2.6. By the Orbit-Stabilizer Theorem, since the action is transitive, the single orbit is the entire set \(A = G \cdot a\):\[|A| = [G : G_a] = \frac{|G|}{|G_a|} = \frac{|G|}{1} = |G|.\]
2.7. Thus \(|G| = |A|\) (the action is free and transitive, i.e., simply transitive / regular).
3. Application to Abelian Transitive Subgroups of \(S_n\):Proof:
3.1. Let \(G \le S_n\) be an abelian transitive subgroup acting on \(A = \{1, 2, \dots, n\}\).
3.2. Since \(G\) is a subgroup of \(S_n = \operatorname{Sym}(A)\), the permutation action of \(G\) on \(A\) is faithful by definition (\(\rho: G \hookrightarrow S_n\) is an inclusion).
3.3. Since \(G\) is abelian and transitive, by Step 2 we have:\[|G| = |A| = n.\]
4. Conclusion:(a) \(\bigcap_{\sigma \in G} \sigma G_a \sigma^{-1} = \ker(\rho) = \{1\}\);(b) \(G_a = \{1\} \implies |G| = |A| = n\). Q.E.D.